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KonstantinChe [14]
3 years ago
14

after a certain pesticide compound is applied to crops its decomposition is a first order reaction with a half life of 56 days.

what is the rate constan for the decomposition reacion
Chemistry
1 answer:
antoniya [11.8K]3 years ago
4 0

Answer:

Rate constant = 0.01238 days⁻¹

Explanation:

The first order reaction has as general equation:

Ln[A] = -kt + Ln[A]₀

<em>Where [A] is concentration of the reactant after time t,</em>

<em>k is rate constant</em>

<em>And [A]₀ the initial concentration of the reactant</em>

<em />

The half-life is the time required to reach the half of the initial concentration. If [A]₀ = 1; [A] = 1/2

Replacing:

Ln[1/2] = -kt + Ln[1]

-0.693 = -k*56days

0.01238days⁻¹ = k

<h3>Rate constant = 0.01238 days⁻¹</h3>

<em />

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Answer:

<h2>Density = 1.67 g/mL</h2>

Explanation:

The density of a substance can be found by using the formula

Density =  \frac{mass}{volume}

From the question

mass = 50 g

volume = 30 mL

Substitute the values into the above formula and solve for the density

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Density =  \frac{50}{30}  \\  =  \frac{5}{3}  \\  = 1.66666...

Wr have the final answer as

<h3>Density = 1.67 g/mL</h3>

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How long does it take for a 12.62g sample of ammonia to heat from 209K to 367K if heated at a constant rate of 6.0kj/min? The me
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First, consider the steps to heat the sample from 209 K to 367K.

1) Heating in liquid state from 209 K to 239.82 K

2) Vaporaizing at 239.82 K

3) Heating in gaseous state from 239.82 K to 367 K.


Second, calculate the amount of heat required for each step.

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=> number of moles = 12.62 g / 17 g/mol = 0.742 mol

Heat1 = #moles * heat capacity * ΔT

Heat1 = 0.742 mol * 80.8 J/mol*K * (239.82K - 209K) = 1,847.77 J

2) Vaporization

Heat2 = # moles * H vap

Heat2 = 0.742 mol * 23.33 kJ/mol = 17.31 kJ = 17310 J

3) Vapor heating

Heat3 = #moles * heat capacity * ΔT

Heat3 = 0.742 mol * 35.06 J / (mol*K) * (367K - 239.82K) = 3,308.53 J

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